Symmetrical Fault Calculations -...
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Symmetrical Fault Calculations Symmetrical Fault Calculations
Transcript of Symmetrical Fault Calculations -...
Symmetrical Fault CalculationsSymmetrical Fault Calculations
Example 1Example 1
34.5kV
SCC = 1200 MVA
10/12.5 MVA6%
13.8 kV
F
Calculate the fault current at F
Example 1Example 1
10/12.5 MVA6%
F
13.8 kV
Calculate the fault current at F
A55.2898kV5.34
MVA100kV
MVAI
ohms9.11100
5.34
MVAkV
Z
5.34kV100MVA
base
basebase
2base
2base
base
base
base
=
==
==
=
==
Step 1: Calculate Base Values34.5kV
SCC = 1200 MVA
Example 1Example 1
10/12.5 MVA6%
F
13.8 kV
Calculate the fault current at F
48.05.12
100x06.0
MVAMVA
xZZoldbase
basetTpu
==
=
Step 2: Calculate p.u impedances34.5kV
SCC = 1200 MVA
Example 1Example 1
10/12.5 MVA6%
F
13.8 kV
Calculate the fault current at F
48.05.12
100x06.0
MVAMVA
xZZoldbase
basetTpu
==
=
Step 2: Calculate p.u impedances
0833.05.345.34
x1200100
kVkV
xMVAMVA
Zbase
oldbase
oldbase
baseutility
=
=
=
34.5kV
SCC = 1200 MVA
Example 1Example 1
10/12.5 MVA6%
F
13.8 kV
Calculate the fault current at F
pu775.10833.048.0
0.1Z
VI
total
puFpu
=+
=
=
Step 3: Calculate fault current34.5kV
SCC = 1200 MVA
Example 1Example 1
10/12.5 MVA6%
F
13.8 kV
Calculate the fault current at F
pu775.10833.048.0
0.1Z
VI
total
puFpu
=+
=
=
Step 3: Calculate fault current
A7426
5.34x8.13
II
A29703
55.2898x775.1
3
IxII
FHVFLV
baseFpuFHV
=
=
==
=
34.5kV
SCC = 1200 MVA
Example 2 Example 2 –– Addition of a GeneratorAddition of a Generator
34.5kV
SCC = 1200 MVA
10/12.5 MVA6% 5MVA
Xd”=0.12
13.8 kV
F
Calculate the fault current at F
Example 2 Example 2 –– Addition of a GeneratorAddition of a Generator
0833.05.345.34
x1200100
kVkV
xMVAMVA
Zbase
oldbase
oldbase
baseutility
=
=
=Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA6% 5MVA
Xd”=0.12
13.8 kV
F
Calculate the fault current at F
Example 2 Example 2 –– Addition of a GeneratorAddition of a Generator
48.05.12
100x06.0
MVAMVA
xZZoldbase
basetTpu
==
=
0833.05.345.34
x1200100
kVkV
xMVAMVA
Zbase
oldbase
oldbase
baseutility
=
=
=Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA6% 5MVA
Xd”=0.12
13.8 kV
F
Calculate the fault current at F
Example 2 Example 2 –– Addition of a GeneratorAddition of a Generator
48.05.12
100x06.0
MVAMVA
xZZoldbase
basetTpu
==
=
0833.05.345.34
x1200100
kVkV
xMVAMVA
Zbase
oldbase
oldbase
baseutility
=
=
=Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA6% 5MVA
Xd”=0.12
13.8 kV
4.25
100x12.0
MVAMVA
xZZoldbase
baseGGpu
==
=F
Calculate the fault current at F
Example 2 Example 2 –– Addition of a GeneratorAddition of a Generator
Step 2: Calculate equivalent impedances34.5kV
SCC = 1200 MVA1.0V
10/12.5 MVA6%
F
13.8 kV
Calculate the fault current at F
5MVAXd
”=0.12
0.0833Zutility ZGpu2.4
ZTpu 0.48
Example 2 Example 2 –– Addition of a GeneratorAddition of a Generator
Step 2: Calculate equivalent impedances
456.04.25633.04.2*5633.0
ZEq =+
=1.0V
0.456
0.0833
Zutility
2.4
ZTpu 0.48
ZGpu
Example 2 Example 2 –– Addition of a GeneratorAddition of a Generator
Step 3: Calculate fault current
456.04.25633.04.2*5633.0
ZEq =+
=1.0V
0.456
pu192.2456.00.1
Z
VI
Eq
puFpu ===
0.0833
Zutility
2.4
ZTpu 0.48
ZGpu
Example 2 Example 2 –– Addition of a GeneratorAddition of a Generator
Step 3: Calculate fault current
456.04.25633.04.2*5633.0
ZEq =+
=1.0V
0.456
pu192.2456.00.1
Z
VI
Eq
puFpu ===
0.0833
Zutility
2.4
ZTpu 0.48
ZGpu
A9170
5.34x8.13
II
A36683
55.2898x192.2
3
IxII
FHVFLV
baseFpuFHV
=
=
==
=
Example 1 & 2 ComparisonExample 1 & 2 Comparison
34.5kV
SCC = 1200 MVA
34.5kV
10/12.5 MVA6%
F
13.8 kV
5MVAXd
”=0.12
IF =9170 Amps
SCC = 1200 MVA
10/12.5 MVA6%
13.8 kV
F
IF =7426 Amps
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
34.5kV
SCC = 1200 MVA
10/12.5 MVA6%
5MVAXd
”=0.12
13.8 kV
F 3 MVA5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
0833.05.345.34
x1200100
kVkV
xMVAMVA
Zbase
oldbase
oldbase
baseutility
=
=
=Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA6%
5MVAXd
”=0.12
13.8 kV
F 3 MVA5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
0833.05.345.34
x1200100
kVkV
xMVAMVA
Zbase
oldbase
oldbase
baseutility
=
=
=
48.05.12
100x06.0
MVAMVA
xZZoldbase
baset1Tpu
==
=
Step 1: Calculate impedances34.5kV
SCC = 1200 MVA
10/12.5 MVA6%
5MVAXd
”=0.12
13.8 kV
F 3 MVA5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
0833.05.345.34
x1200100
kVkV
xMVAMVA
Zbase
oldbase
oldbase
baseutility
=
=
=
48.05.12
100x06.0
MVAMVA
xZZoldbase
baset1Tpu
==
=
Step 1: Calculate impedances34.5kV
SCC = 1200 MVA
10/12.5 MVA6%
5MVAXd
”=0.12
13.8 kV
F 3 MVA5%
2.4 kV
4.25
100x12.0
MVAMVA
xZZoldbase
baseGGpu
==
=M 2500 HP
Xd”=0.17
Calculate the fault current at F
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
pu4.2Z
pu48.0Z
pu0833.0Z
Gpu
1Tpu
utility
=
=
=Step 1: Calculate impedances34.5kV
SCC = 1200 MVA
10/12.5 MVA6%
5MVAXd
”=0.12
F
13.8 kV
Calculate the fault current at F
667.13
100x05.0
MVAMVA
xZZoldbase
baset2Tpu
==
=
3 MVA5%
2.4 kV
M 2500 HPXd
”=0.17
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
pu667.1Z
pu4.2Z
pu48.0Z
pu0833.0Z
2Tpu
Gpu
1Tpu
utility
=
=
=
=Step 1: Calculate impedances34.5kV
SCC = 1200 MVA
10/12.5 MVA6%
5MVAXd
”=0.12
89.7154.2
100x17.0
MVAMVA
xZZ
2154866.0
1865pf
746.0*2500ratingMotor
oldbase
basemMpu
==
=
==
=13.8 kV
F 3 MVA5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
Step 1: Calculate impedances34.5kV
SCC = 1200 MVA
pu89.7Z
pu667.1Z
pu4.2Z
pu48.0Z
pu0833.0Z
Mpu
2Tpu
Gpu
1Tpu
utility
=
=
=
=
=10/12.5 MVA6%
5MVAXd
”=0.12
13.8 kV
F 3 MVA5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
Step 2: Calculate Equivalent impedances34.5kV
SCC = 1200 MVA1.0V
10/12.5 MVA6%
5MVAXd
”=0.12
F
13.8 kV
Calculate the fault current at F
3 MVA5%
2.4 kV
M 2500 HPXd
”=0.17
ZTpu1
0.0833
Zutility
0.48
ZGpu
2.4
7.89
1.667
ZMpu
ZTpu2
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
Step 2: Calculate Equivalent impedances
1.0V 1.0V
ZTpu1
0.0833
Zutility
0.48
ZGpu
2.4
7.89
1.667
ZMpu 0.4354
ZTpu2
Example 3 Example 3 ––Presence of transformer & motorPresence of transformer & motor
Step 3: Calculate fault current
pu297.24354.0
0.1Z
VI
Eq
puFpu ===
1.0V
A9609
5.34x8.13
II
A38443
55.2898x297.2
3
IxII
FHVFLV
baseFpuFHV
=
=
==
=
0.4354
Symmetrical FaultSymmetrical Fault
34.5kV34.5kV34.5kV SCC = 1200 MVA
SCC = 1200 MVASCC = 1200 MVA
10/12.5 MVA6%
10/12.5 MVA6%
10/12.5 MVA6%
5MVAXd
”=0.12
13.8 kV5MVAXd
”=0.12
F
IF =7426 Amps
F
13.8 kV
F
13.8 kV
3 MVA5%
2.4 kV
M 2500 HPXd
”=0.17
IF =9170 Amps IF =9609 Amps
